The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(X, Y)

Rewrite Strategy: FULL

(1) RenamingProof (EQUIVALENT transformation)

Renamed function symbols to avoid clashes with predefined symbol.

(2) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0', g(X, Y))
g(0', Y) → 0'
g(X, s(Y)) → g(X, Y)

S is empty.
Rewrite Strategy: FULL

(3) TypeInferenceProof (BOTH BOUNDS(ID, ID) transformation)

Infered types.

(4) Obligation:

TRS:
Rules:
h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0', g(X, Y))
g(0', Y) → 0'
g(X, s(Y)) → g(X, Y)

Types:
h :: s:0' → s:0' → h:f
f :: s:0' → s:0' → s:0' → h:f
s :: s:0' → s:0'
g :: s:0' → s:0' → s:0'
0' :: s:0'
hole_h:f1_0 :: h:f
hole_s:0'2_0 :: s:0'
gen_s:0'3_0 :: Nat → s:0'

(5) OrderProof (LOWER BOUND(ID) transformation)

Heuristically decided to analyse the following defined symbols:
h, g

They will be analysed ascendingly in the following order:
g < h

(6) Obligation:

TRS:
Rules:
h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0', g(X, Y))
g(0', Y) → 0'
g(X, s(Y)) → g(X, Y)

Types:
h :: s:0' → s:0' → h:f
f :: s:0' → s:0' → s:0' → h:f
s :: s:0' → s:0'
g :: s:0' → s:0' → s:0'
0' :: s:0'
hole_h:f1_0 :: h:f
hole_s:0'2_0 :: s:0'
gen_s:0'3_0 :: Nat → s:0'

Generator Equations:
gen_s:0'3_0(0) ⇔ 0'
gen_s:0'3_0(+(x, 1)) ⇔ s(gen_s:0'3_0(x))

The following defined symbols remain to be analysed:
g, h

They will be analysed ascendingly in the following order:
g < h

(7) RewriteLemmaProof (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
g(gen_s:0'3_0(a), gen_s:0'3_0(+(1, n5_0))) → *4_0, rt ∈ Ω(n5₀)

Induction Base:
g(gen_s:0'3_0(a), gen_s:0'3_0(+(1, 0)))

Induction Step:
g(gen_s:0'3_0(a), gen_s:0'3_0(+(1, +(n5_0, 1)))) →_R^Ω(1)
g(gen_s:0'3_0(a), gen_s:0'3_0(+(1, n5_0))) →_IH
*4_0

We have rt ∈ Ω(n¹) and sz ∈ O(n). Thus, we have irc_R ∈ Ω(n).

(8) Complex Obligation (BEST)

(9) Obligation:

TRS:
Rules:
h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0', g(X, Y))
g(0', Y) → 0'
g(X, s(Y)) → g(X, Y)

Types:
h :: s:0' → s:0' → h:f
f :: s:0' → s:0' → s:0' → h:f
s :: s:0' → s:0'
g :: s:0' → s:0' → s:0'
0' :: s:0'
hole_h:f1_0 :: h:f
hole_s:0'2_0 :: s:0'
gen_s:0'3_0 :: Nat → s:0'

Lemmas:
g(gen_s:0'3_0(a), gen_s:0'3_0(+(1, n5_0))) → *4_0, rt ∈ Ω(n5₀)

Generator Equations:
gen_s:0'3_0(0) ⇔ 0'
gen_s:0'3_0(+(x, 1)) ⇔ s(gen_s:0'3_0(x))

The following defined symbols remain to be analysed:
h

(10) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)

Could not prove a rewrite lemma for the defined symbol h.

(11) Obligation:

TRS:
Rules:
h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0', g(X, Y))
g(0', Y) → 0'
g(X, s(Y)) → g(X, Y)

Types:
h :: s:0' → s:0' → h:f
f :: s:0' → s:0' → s:0' → h:f
s :: s:0' → s:0'
g :: s:0' → s:0' → s:0'
0' :: s:0'
hole_h:f1_0 :: h:f
hole_s:0'2_0 :: s:0'
gen_s:0'3_0 :: Nat → s:0'

Lemmas:
g(gen_s:0'3_0(a), gen_s:0'3_0(+(1, n5_0))) → *4_0, rt ∈ Ω(n5₀)

Generator Equations:
gen_s:0'3_0(0) ⇔ 0'
gen_s:0'3_0(+(x, 1)) ⇔ s(gen_s:0'3_0(x))

No more defined symbols left to analyse.

(12) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
g(gen_s:0'3_0(a), gen_s:0'3_0(+(1, n5_0))) → *4_0, rt ∈ Ω(n5₀)

(13) BOUNDS(n^1, INF)

(14) Obligation:

TRS:
Rules:
h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0', g(X, Y))
g(0', Y) → 0'
g(X, s(Y)) → g(X, Y)

Types:
h :: s:0' → s:0' → h:f
f :: s:0' → s:0' → s:0' → h:f
s :: s:0' → s:0'
g :: s:0' → s:0' → s:0'
0' :: s:0'
hole_h:f1_0 :: h:f
hole_s:0'2_0 :: s:0'
gen_s:0'3_0 :: Nat → s:0'

Lemmas:
g(gen_s:0'3_0(a), gen_s:0'3_0(+(1, n5_0))) → *4_0, rt ∈ Ω(n5₀)

Generator Equations:
gen_s:0'3_0(0) ⇔ 0'
gen_s:0'3_0(+(x, 1)) ⇔ s(gen_s:0'3_0(x))

No more defined symbols left to analyse.

(15) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
g(gen_s:0'3_0(a), gen_s:0'3_0(+(1, n5_0))) → *4_0, rt ∈ Ω(n5₀)

(16) BOUNDS(n^1, INF)